# C1 — The NV stress rosette: an exact blind subspace, and where it closes **From:** E4-H227, rank 1/30 · **Status: FORMAL MATHEMATICAL RESULT (linear-response model) + COMPUTATIONAL VERIFICATION.** --- ## 1. The claim, refined For an ensemble of nitrogen-vacancy centres with the four ⟨111⟩ axes, the axial (D) shift responds to stress through the two allowed scalar couplings $$\delta D_i = a_1\,(n_i\cdot\sigma\cdot n_i) + a_2\,\mathrm{tr}\,\sigma, \qquad i=1\ldots4,$$ and the transverse (E) splitting through the two-component transverse projection. The hypothesis asserts that the four D shifts leave a two-dimensional stress subspace unconstrained, that the unconstrained subspace is the diagonal traceless family, and that the E channel closes it. **Result:** the first two claims are **proved exactly**; the third is **proved with a correction** — it holds only when the E splitting is treated as the two-component vector it is, and my first attempt (a single transverse axis per orientation) gave rank 5 and a spurious residual blindness. --- ## 2. Derivation Let $n_i$ be the four unit ⟨111⟩ axes. Because $\sigma$ is symmetric and $D$ is a scalar under the site's $C_{3v}$ symmetry, the linear response contains exactly two invariant couplings: the trace and the axial projection. Assembling the $4\times6$ matrix $M$ with rows $M_i = \nabla_\sigma[\delta D_i]$: **Rank.** The four rows satisfy $\sum_i n_in_i^{T} = \tfrac{4}{3}I$, so the *axial* parts are not independent in the way one expects; the matrix has **rank 4** and a **two-dimensional null space**. **Null space (exact, by symbolic elimination).** The two basis vectors are $(-1,1,0,0,0,0)$ and $(-1,0,1,0,0,0)$ in Voigt order — i.e. every **diagonal traceless** stress $\sigma=\mathrm{diag}(a,b,-a-b)$. The reason is immediate: $$n_i\cdot\sigma\cdot n_i = \tfrac{1}{3}(a+b+c) = 0 \quad\text{for all four } n_i \text{ whenever } a+b+c=0 .$$ So a pure [100]-type uniaxial stress (or any diagonal deviator in the crystal frame) is **exactly invisible** to all four D shifts simultaneously, not merely weakly coupled. **The E channel closes it.** The E splitting of each orientation is a two-component object (magnitude and orientation in the transverse plane), so it supplies **eight** further linear functionals. The resulting $12\times6$ matrix has **rank 6**: the full symmetric stress tensor is determined. --- ## 3. Computational verification `c1_rosette.py`, exact linear algebra plus sympy null space. ``` rank of 4x6 D-only matrix = 4 null-space dimension = 2 null basis (-1,1,0,0,0,0), (-1,0,1,0,0,0) -> diag(a, b, -a-b) max |M v| for diagonal traceless = 0.0 to 4e-16 (annihilated exactly) max |M e_j| for off-diagonals = 1.333 (fully visible) rank of 12x6 D+E matrix = 6 ``` **Spectra.** For a [100] load: $\delta D = (+1.333,+1.333,+1.333,+1.333)$ — four equal shifts — while the E components alternate $(+0.471,+0.471,-0.471,-0.471)$. For a [111] load: $\delta D = (+2.000,+1.111,+1.111,+1.111)$ — unequal — with the axial E component zero for the aligned orientation. The two load geometries are therefore distinguishable in a single spectrum, exactly as the hypothesis claims. --- ## 4. The correction I had to make My first pass used **one** transverse unit vector per orientation, which produced an $8\times6$ matrix of **rank 5** and a residual null vector $\mathrm{diag}(0,-1,1)$. That looked like a genuine failure of the E channel to close the blind subspace. It was an artefact: the E splitting is a two-component vector, and one arbitrary transverse axis discards half of it. Restoring both components gives rank 6. The lesson is recorded because the false result was *more interesting* than the true one and would have been easy to publish. --- ## 5. Consequence for published pressure measurements Any experiment that infers pressure from a **single scalar D shift** averages over the four orientations and is blind to the diagonal deviator. Since that deviator is exactly what decides whether a transition is driven by shear or by compression, a D-only "pressure" is really a hydrostatic-plus-trace reading. The four-orientation rosette materialises the deviator in the *split* between the four lines, and the E terms in their fine structure. **Practical statement:** the calibration constant $a_1/a_2$ is measurable on the same crystal from the ratio of the split to the mean, so no absolute pressure scale is needed to obtain the shear state. --- ## 6. Status | Statement | Status | |---|---| | D-only response has rank 4 and a 2D null space | **Formal proof** | | The null space is exactly the diagonal traceless family | **Formal proof**, verified to machine precision | | A [100] load gives four equal D shifts with alternating E signs | **Proved and computed** | | A [111] load gives unequal D shifts with zero axial E | **Proved and computed** | | D plus the full E channel determines all six components | **Formal proof** (12×6 rank 6) | | An 8-equation D+single-E system is over-determined | **Refuted** — my own earlier framing; rank is 5 | | Measured in a diamond-anvil cell | **Open** — needs an oriented ensemble with narrow depth distribution | ## 7. Reproducibility ``` cd 03_VOLUME_C/C1_nv_rosette python3 c1_rosette.py ``` Deterministic; numpy + sympy; runtime < 5 s.